Young's Double Slit Experiment is famous because it demonstrated that light, theretofore thought of as solely discrete particles called corpuscles, actually had properties of waves in addition to particles. This was demonstrated by the two light beams separated by the note card used by Young constructively and destrucively interfering with each other to form bands of light and darkness, which showed that light acts like a wave. However, the light that reached the wall was made up of discrete points, which could only occur if light also could behave like it was made of particles. Therefore, it was concluded that light can act both like a particle and a wave, known as wave-particle duality.

This question is just a quick application of the double slit formula. 

, where  is the distance the  bright fringe is from the center of the screen.  is the wavelength,  is the distance the slit is from the center of the screen, and  is the slit width. If you set , the value of  you get is also the distance between each adjacent bright fringe (the first and second are adjacent).

Note-The value we got is very large because of the very small slit width.

In an electromagnetic wave, the electric and magnetic portions are proprtional to each other. The ratio of electric field to magnetic field is , the speed of light.

Using this information, we can determine the strength of one part given the other.

We have  and , so now we multiply the two.

Therefore, the strength of the electric field is .

In an electromagnetic wave going through a vacuum, the ratio of  and  is the speed of light, ; however, this relation doesn't change when light goes through a medium. The ratio still is the speed of light, but instead of , it's the speed of light through the medium. Therefore, for velocity , 

In this problem, we're told that the speed of light through the medium is , so if we rearrange the equation to solve for  and use  in place of v, we'll get our answer.

Therefore, the strength of the magnetic field is .

Wavelength is indirectly related to the amount of energy in the wave (the frequency). The most energetic wves are gamma rays, while the least energetic are radio waves. This means that radio waves have the longest wavelength.

A helpful mnemonic for remembering the order of wavelengths in the electromagnetic spectrum from longest to shortest is "Raging Martians invaded Roy G Biv using x-ray guns."

In order, the letters stand for 

1. Radio
2. Micro
3. Infrared
4. Red
5. Orange
6. Yellow
7. Green
8. Blue
9. Indigo
10. Violet
11. Ultraviolet
12. X-ray
13. Gamma

When light is refracted, the average speed is lower if the index of refraction is higher (and vice versa). In order to refract, the medium light travels through must have a changing index of refraction. Because the amount of energy the light carries must be constant during refraction, the frequency (which is proportional to energy) cannot change. Therefore, since  and the frequency must stay the same when the speed changes, wavelength must also change.

When trying to find the minimum distance between two objects that makes them still distinguishable, we use the Rayleigh criterion equation.

 is the minimum angle at which two objects can be resolved,  is the wavelength of the wave being used, and  is the diameter of the circular aperture (in this case, the antenna would be the aperture). An important part of equations like these is the small-angle approximation, which states, as the angle approaches zero,

This means that, if the angle is small enough, . In our problem, we're trying to find , so we can substitute  for 

Next, since we're not given the wavelength, but we are given the frequency, we can find the wavelength using the following equation: 

Now, solve for .

Using

Where  is the speed of light, is frequency and is wavelength

Converting to and plugging in values

Solving for

For this question, we're given the distance between wave peaks (wavelength) as well as how long it takes for each wave to hit the shore (period). We're then asked to use this information to solve for the speed of the waves.

First, we'll need to use an equation for the speed of a wave.

We know from the question stem that we have the wavelength, which is the distance between the peaks of each wave. But we do not have the frequency. However, we can solve for frequency by using the period of the wave as follows:

Which we can alternatively write as:

The period of the wave, , represents the amount of time it takes for a single wave to pass a given point, and thus has units of time. The inverse of this quantity is the frequency, which describes how many waves pass by a given point in a certain amount of time. Usually, the unit of time in these cases is seconds.

Thus, for every one second that passes, one-fifth of a wave will pass by any given point.

Now, we can plug this value into the equation for the speed of the wave to solve for our answer:

To answer this question, let's first recall that on the visible spectrum, red light has the longest wavelength. Due to this, red light also has the lowest energy. We can show this with the equation

As the dimmer is gradually turned down, less and less energy is being provided to the light bulb. Consequently, since red light has the least amount of energy of all the colors on the visible spectrum, we briefly see red light the instant before the light turns off completely. Note that the speed of a wave is determined by the medium. In this case, regardless of the wavelength, the medium is air, which determines the speed of any electromagnetic wave.